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In Brian Hall's Lie Groups, Lie Algebras, and Representations, he defines the dual basis of a inner product space $E$ as follows:

For any basis $\{v_1, \cdots, v_n\}$ of $E$, we can form a basis for the dual space $E^*$ by considering the linear functionals $\xi_j$ given by $\xi_j(v_k) = \delta_{jk}$. We can then find unique vectors $v_j^*$ of $E$ such that $\xi_j(u) = \langle v_j^*, u \rangle$ so that $\langle v_j^*, v_k \rangle = \delta_{jk}$. The basis $\{v_1^*, \cdots, v_n^*\}$ for $E$ is called the dual basis to $\{v_1, \cdots, v_n\}$.

I honestly got quite confused with his definition, because the usual definition of a dual basis is the $\xi_j$'s on $E^*$ and is never meant to be a basis for $E$. And I can't seem to find any reference that uses this definition of dual basis. So here is a few questions:

(1) How do I find these $v_j^*$'s (perhaps without using the dual space).

(2) What would be the geometric relationship between the original basis and the dual basis, say for $E = \mathbb{R}^n$? (I guess this will also give an alternative answer to (1).)

(3) What would the change-of-basis matrix look like?

He also claims (Lemma 8.41) that if the original basis is obtuse, then the dual basis is acute, and I just can't even think of a linear transofrmation that does this job.

Ray
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  • There is a typo in the second line of the definition. Replace $E$ by $E^{*}$. – Kavi Rama Murthy Dec 25 '23 at 08:08
  • Uh ... nope, that's correct. The definition is carefully copied word by word and I proofread it multiple times. He does use $v_j^*$ for elements in $E$ (which is also why I'm so confused with his definition) – Ray Dec 25 '23 at 08:12
  • Typo is not by you. It is a typo in the book itself. – Kavi Rama Murthy Dec 25 '23 at 08:19
  • In an inner product space, the dual basis certainly should be a basis of the vector space itself. The book is correct. – Claudius Dec 25 '23 at 08:24
  • Humm, maybe, but then the equation $\langle v_j^, v_k \rangle = \delta_{jk}$ would mean $\langle -,- \rangle$ is a function $E^ \times E \to \mathbb{R}$, which I think many people uses to mean "evaluate the function at the second input", but I think Brian never meant (and said) to use it this way (and thus an inner product should be a symmetric positive-definite bilinear map $E \times E \to \mathbb{R}$). – Ray Dec 25 '23 at 08:33
  • An inner product is a certain map $E\times E \to \mathbb{F}$, and $v_j^$ and $v_k$ lie in $E$. So why do you conclude that the inner product must be a function $E^\times E \to \mathbb{F}$? – Claudius Dec 25 '23 at 08:35
  • What I meant is in the case where it is a typo: if it is a typo and $v_j^$ is meant to be in $E^$ then the usage $\langle v_j^, v_k \rangle$ would mean $\langle -,- \rangle$ has to be a function $E^ \times E \to \mathbb{F}$. – Ray Dec 25 '23 at 09:23
  • I cannot find Lemma 8.41 in Hall's book. Can you give more details (e.g. statement of the lemma and necessary definitions)? – Claudius Dec 25 '23 at 13:36
  • In $\mathbb{R}^n$, $v_j^* = \xi_j^T$. – Mason Dec 25 '23 at 15:46

2 Answers2

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The book is not unique in using duality terminology that way. In fact some classic books say that two vector spaces $V',V$ over a field $\mathbb{F}$ are "dual" whenever there exists a non-degenerate bilinear form (i.e. pairing) $\langle-,-\rangle:V'\times V\to\mathbb{F}$. Here $V'$ is not assumed to be the space of linear forms on $V$.

When using this definition, if $E$ is a real inner product space, then $E$ is self-dual via the inner product $\langle-,-\rangle:E\times E\to\mathbb{R}$. If $E$ is a complex inner product space, then $E,\overline{E}$ are dual via the inner product where $\overline{E}$ is the conjugate space of $E$ with the same underlying additive group but scalar multiplication defined using complex conjugation. In what follows I'll assume $E$ is a real inner product space to keep things simpler.

To find the dual basis vectors $v_i^*$, we can write $$v_i^*=\sum_k\alpha_{ik}v_k\tag{1}$$ Taking an inner product with $v_j$ on both sides of (1), we obtain $$\delta_{ij}=\xi_i(v_j)=\langle v_i^*,v_j\rangle=\sum_k\alpha_{ik}\langle v_k,v_j\rangle\tag{2}$$ Writing $I$ for the identity matrix, $A=(\alpha_{ij})$ for the coefficient matrix, and $G=(\langle v_i,v_j\rangle)$ for the Gram matrix (metric tensor) of the original basis, (2) just says $I=AG$, so $A=G^{-1}$.

Taking an inner product with $v_j^*$ on both sides of (1), we also obtain $$\langle v_i^*,v_j^*\rangle=\sum_k\alpha_{ik}\delta_{jk}=\alpha_{ij}\tag{3}$$ so $A$ can be interpreted as the Gram matrix of the dual basis vectors. If we port the inner product space structure to $E^*$ using the Riesz isomorphism, $x\mapsto\langle x,-\rangle$, then $A$ is also the Gram matrix of the $\xi_i$. The change of basis matrix is obtained from $A$.

Notice that the basis $v_1,\ldots,v_n$ is orthonormal precisely when it is self-dual, in which case $A=G=I$.

blargoner
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Suppose $E = \mathbb{R}^n$. The Riesz isomorphism identifies $v^T \in (\mathbb{R}^n)^*$ with $v \in \mathbb{R}^n$. We have $\xi_j \cdot v_k = \delta_{jk}$, so $\Xi^T V = I$, where $\Xi$ has columns $\xi_1, \dots, \xi_n$ and $V$ has columns $v_1, \dots, v_n$. Thus $\Xi = (V^T)^{-1}$. Thus the dual basis is the columns of $(V^T)^{-1}$.

Mason
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